Group together arbitrary date objects that are within a time range of each other

Question

I want to split the calendar into two-week intervals starting at 2008-May-5, or any arbitrary starting point.

So I start with several date objects:

import datetime as DT

raw = ("2010-08-01",
       "2010-06-25",
       "2010-07-01",
       "2010-07-08")

transactions = [(DT.datetime.strptime(datestring, "%Y-%m-%d").date(),
                 "Some data here") for datestring in raw]
transactions.sort()

By manually analyzing the dates, I am quite able to figure out which dates fall within the same fortnight interval. I want to get grouping that's similar to this one:

# Fortnight interval 1
(datetime.date(2010, 6, 25), 'Some data here')
(datetime.date(2010, 7, 1), 'Some data here')
(datetime.date(2010, 7, 8), 'Some data here')

# Fortnight interval 2
(datetime.date(2010, 8, 1), 'Some data here')

Answer 1

import datetime as DT
import itertools

start_date=DT.date(2008,5,5)

def mkdate(datestring):
    return DT.datetime.strptime(datestring, "%Y-%m-%d").date()

def fortnight(date):
    return (date-start_date).days //14

raw = ("2010-08-01",
       "2010-06-25",
       "2010-07-01",
       "2010-07-08")
transactions=[(date,"Some data") for date in map(mkdate,raw)]
transactions.sort(key=lambda (date,data):date)

for key,grp in itertools.groupby(transactions,key=lambda (date,data):fortnight(date)):
    print(key,list(grp))

yields

# (55, [(datetime.date(2010, 6, 25), 'Some data')])
# (56, [(datetime.date(2010, 7, 1), 'Some data'), (datetime.date(2010, 7, 8), 'Some data')])
# (58, [(datetime.date(2010, 8, 1), 'Some data')])

Note that 2010-6-25 is in the 55th fortnight from 2008-5-5, while 2010-7-1 is in the 56th. If you want them grouped together, simply change start_date (to something like 2008-5-16).

PS. The key tool used above is itertools.groupby, which is explained in detail here.

Edit: The lambdas are simply a way to make "anonymous" functions. (They are anonymous in the sense that they are not given names like functions defined by def). Anywhere you see a lambda, it is also possible to use a def to create an equivalent function. For example, you could do this:

import operator
transactions.sort(key=operator.itemgetter(0))

def transaction_fortnight(transaction):
    date,data=transaction
    return fortnight(date)

for key,grp in itertools.groupby(transactions,key=transaction_fortnight):
    print(key,list(grp))

Answer 2

Use itertools groupby with lambda function to divide by the length of period the distance from starting point.

>>> for i, group in groupby(range(30), lambda x: x // 7):
    print list(group)


[0, 1, 2, 3, 4, 5, 6]
[7, 8, 9, 10, 11, 12, 13]
[14, 15, 16, 17, 18, 19, 20]
[21, 22, 23, 24, 25, 26, 27]
[28, 29]

So with dates:

import itertools as it
start = DT.date(2008,5,5)
lenperiod = 14

for fnight,info in it.groupby(transactions,lambda data: (data[0]-start).days // lenperiod):
    print list(info)

You can use also weeknumbers from strftime, and lenperiod in number of weeks:

for fnight,info in it.groupby(transactions,lambda data: int (data[0].strftime('%W')) // lenperiod):
    print list(info)